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Mathematics Question on Geometrical Meaning of the Zeroes of a Polynomial

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2–2x–8x^2 – 2x – 8 (ii) 4s2–4s+14s^2 – 4s + 1 (iii) 6x2–3–7x6x^2 – 3 – 7x (iv) 4u2+8u4u^2 + 8u (v) t2–15 t^2 – 15 (vi) 3x2–x–43x^2 – x – 4

Answer

**(i) **x2−2x−8x^2-2x-8
=(x−4)(x+2)=(x-4)(x+2)

The value of x2−2x−8x^2-2x-8 is zero when x−4=0x - 4 = 0 or x+2=0x + 2 = 0 ,

i.e., when x=4x = 4 or x=−2x = -2
Therefore, the zeroes of x2−2x−8x^2-2x-8 are 44 and −2.-2.
Sum of zeroes =4−2=2== 4-2=2= −(−2)1\dfrac{-(-2)}{1} =−(-(Coefficient of x)x) Coefficient of x2x^2
Product of zeroes =4x(−2)=−8=(−8)1= Constant termCoefficient of x2= 4x(-2)=-8= \dfrac{(-8)}{1} = \dfrac{\text{ Constant term}}{\text{Coefficient of }x^2}

(ii) 4s2−4s+1 4s^2-4s+1
=(2s−1)2=(2s-1)^2

The value of 4s2−4s+1 4s^2-4s+1 is zero when 2s−1=0,2s - 1 = 0,

i.e.,s=12s = \dfrac{1}{2} Therefore, the zeros of 4s2−4s+14s^2 - 4s + 1 are 12\dfrac{1}{2} and 12.\dfrac{1}{2} .
Therefore,
The sum of zeroes=12+12=11=−(−4)4 = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{1}{1}= \dfrac{-(-4)}{ 4} [Multiply by 44 in Numerator and Denominator]=−(Coefficient of s)(Coefficient of s2)= \dfrac{-\text{(Coefficient of s)}}{\text{(Coefficient of } s^2)}
Product of zeroes =12×12=14= Constant termCoefficient of S2= \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} = \dfrac{\text{ Constant term}}{\text{Coefficient of }S^2}

**(iii) **6x2−3−7x6x^2-3-7x
=6x2−7x−3=(3x+1)(2x−3)= 6x^2-7x-3=(3x+1)(2x-3)

The value of 6x2−3−7x6x^2 - 3 - 7x is zero when 3x+1=03x + 1 = 0 or 2x−3=02x -3 = 0, i.e.,
x=−13x = \dfrac{-1}{3} or x=32x = \dfrac{3}{2}

Therefore, the zeroes of 6x2−3−7x6 x^2 − 3 − 7 x are 6x2−3−3−7x6x^2 -3 -3 -7x are −13\dfrac{-1}{3} and 32\dfrac{3}{2}.

Sum of Zeros =−13+32=76=−(−7)6=−(Coefficient of x)Coefficient of x2= \dfrac{-1}{3} + \dfrac{3}{2} = \dfrac{7}{6} = \dfrac{-(-7)}{6} = \dfrac{-(\text{Coefficient of } x)}{ \text{Coefficient of } x^2}

Product of zeroes =−13×32=−12=−36= Constant termCoefficient of x2= \dfrac{-1}{3} \times \dfrac{3}{2} = \dfrac{-1}{ 2} = \dfrac{-3}{6} = \dfrac{\text{ Constant term}}{\text{Coefficient of }x^2}

(iv) 4u2+8u4u^2 + 8u
=4u2+8u+0= 4u^2 + 8u +0
=4u(u+2)=4u(u+2 )

The value of 4u2+8u4u^2 + 8u is zero when 4u=04u = 0 or u+2=0u + 2 = 0, i.e.,
u=0u = 0 or u=−2u = −2

Therefore, the zeroes of 4u2+8u4u^2 + 8u are 00 and −2.−2.

The sum of Zeros =0+(−2)=−2=−(8)4=0 + (-2) = -2 = \dfrac{-(8)}{4} [Multiply by 44 in Numerator and Denominator]=−(Coefficient of u)(Coefficient of u2)= \dfrac{-\text{(Coefficient of u)}}{\text{(Coefficient of } u^2)}
Product of zeroes = 0 \times (-2) =0 = \dfrac{0}{4} = $$\dfrac{\text{ Constant term}}{\text{Coefficient of }u^2}

**(v) **t2−15t^2 -15
=t2−0.t−15= t^2 -0.t -15
=(t−15)(t+15)=(t-\sqrt{15}) (t + \sqrt {15})

The value of t2−15t^2 − 15 is zero when t−15=0t-\sqrt{15} = 0 or t+15=0t + \sqrt{15} =0,

i.e., whent=15t = \sqrt{15} ort=−15t = -\sqrt{15}

Therefore, the zeroes of t2−15t^2 − 15 are15 and −15.\sqrt{15} \text{ and } -\sqrt{15}.

Sum of Zeros= \sqrt{15} + (-\sqrt{15}) = 0 =-0/1 = -$$\dfrac{-\text{(Coefficient of t)}}{\text{(Coefficient of } t^2)}

Product of zeroes =15×(−sqrt15)=−15=−15/1== \sqrt{15} \times (-sqrt{15}) = -15 = -15/1=  Constant termCoefficient of t2\dfrac{\text{ Constant term}}{\text{Coefficient of }t^2}

**(vi) **3x2−x−43x^2 − x − 4
=(3x−4)(x+1)= (3x-4)(x+1)

The value of 3x2−x−43x^2 − x − 4 is zero when 3x−43x − 4 =0= 0 or x+1=0x + 1 = 0,

i.e., whenx=43 or x=−1x= \dfrac{4}{3} \text{ or } x = −1

Therefore, the zeroes of 3x2−x−43x^2 − x − 4 are 43\dfrac{4}{3} and −1−1.

Sum of Zeros = \dfrac{4}{3} + (-1) = \dfrac{1}{3} = \dfrac{-(-1)}{3} =$$ \dfrac{-\text{(Coefficient of x)}}{\text{(Coefficient of } x^2)}
Product of zeroes =43×(−1)= \dfrac{4}{3} \times (-1) = =43= \dfrac{4}{3}=  Constant termCoefficient of x2\dfrac{\text{ Constant term}}{\text{Coefficient of }x^2}